文科高数题目lim[1/(x+1)-3/(x^3+1)] x趋向于-1 lim[(e^2x-1)/x] x趋向于0 没学过罗比达法则..
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文科高数题目lim[1/(x+1)-3/(x^3+1)] x趋向于-1 lim[(e^2x-1)/x] x趋向于0 没学过罗比达法则..
文科高数题目
lim[1/(x+1)-3/(x^3+1)] x趋向于-1
lim[(e^2x-1)/x] x趋向于0
没学过罗比达法则..
文科高数题目lim[1/(x+1)-3/(x^3+1)] x趋向于-1 lim[(e^2x-1)/x] x趋向于0 没学过罗比达法则..
1/(x+1)-3/(x^3+1)
=[(x^2-x+1)-3]/(x^3+1)
=(x^2-x-2)/(x^3+1)
=(x-2)(x+1)/(x+1)(x^2-x+1)
=(x-2)/(x^2-x+1)
所以lim(x→-1)[1/(x+1)-3/(x^3+1)]
=lim(x→-1)(x-2)/(x^2-x+1)
=(-1-2)/(1+1+1)
=-1
当x趋向于0时,e^(2x)-1和x趋于0
是0/0型,可以用洛必达法则
lim(x→0)[(e^2x-1)/x]
=lim(x→0)2e^2x
=2
[1/(x+1)-3/(x^3+1)] =(x^2-x+1)/(x^3+1)-3/(x^3+1)
=(x^2-x-2)/(x^3+1)
=(x-2)(x+1)/(x+1)(x^2-x+1)
=(x-2)/(x^2-x+1)
代入-1
所以为-3/3=-1
(1)属于无穷大-无穷大型,先通分,变成0/0型,然后采用罗比达法则,得到结果-1.
(2)属于0/0型,最简单的办法是对分子采用等价无穷小代换:当x趋于0时,e^x等价于x;从而分子部分等价于2x,然后跟分母一约去x即得2.此题也可以直接运用罗比达法则,通过对分子分母分别求导数,结果也是2....
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(1)属于无穷大-无穷大型,先通分,变成0/0型,然后采用罗比达法则,得到结果-1.
(2)属于0/0型,最简单的办法是对分子采用等价无穷小代换:当x趋于0时,e^x等价于x;从而分子部分等价于2x,然后跟分母一约去x即得2.此题也可以直接运用罗比达法则,通过对分子分母分别求导数,结果也是2.
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lim[1/(x+1)-3/(x^3+1)]
x^3+1=(x+1)(x^2-x+1)
原式=lim[(x^2-x+1-3)/(x^3+1)]=lim[(x-2)(x+1)/(x^3+1)
=lim[(x-2)/(x^2-x+1)]
当x趋向-1
原式=[(-1-3)/(1+1+1)]=-1
lim[1/(x+1)-3/(x^3+1)]=lim{1/(x+1)-3/[(x+1)(x^2-x+1)]}
=lim{(x^2-x+1)/[(x+1)(x^2-x+1)]-3/[(x+1)(x^2-x+1)]}
=lim{(x^2-x-2)/[(x+1)(x^2-x+1)]}
=lim{(x-2)(x+1)/[(x+1)(x^2-x+1)]}
=lim[(x-2)...
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lim[1/(x+1)-3/(x^3+1)]=lim{1/(x+1)-3/[(x+1)(x^2-x+1)]}
=lim{(x^2-x+1)/[(x+1)(x^2-x+1)]-3/[(x+1)(x^2-x+1)]}
=lim{(x^2-x-2)/[(x+1)(x^2-x+1)]}
=lim{(x-2)(x+1)/[(x+1)(x^2-x+1)]}
=lim[(x-2)/(x^2-x+1)]
当x→-1,lim[(x-2)/(x^2-x+1)]→lim(-3/3)=-1
lim[(e^2x-1)/x] 不会做,工作时间长,很久没做高数题目了呵
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